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Sakurai page 91: Simple Harmonic Oscillator, trouble understanding

  1. May 1, 2013 #1
    From page 91 of "Modern Quantum Mechanics, revised edition", by J. J. Sakurai.

    Some operators used below are,
    [tex]
    a = \sqrt{\frac{m \omega}{2 \hbar}} \left(x + \frac{ip}{m \omega} \right)\\
    a^{\dagger} = \sqrt{\frac{m \omega}{2 \hbar}} \left(x - \frac{ip}{m \omega} \right)\\
    N = a^{\dagger} a
    [/tex]

    In equation (2.3.16), he finds
    [tex]
    a \mid n \rangle = \sqrt{n} \mid n - 1 \rangle
    [/tex]
    Next he says it's similarly easy to show equation (2.3.17),
    [tex]
    a^{\dagger} \mid n \rangle = \sqrt{n+1} \mid n + 1 \rangle
    [/tex]
    I didn't find it to be so easy. Following the same procedure as equation (2.3.16), since from equation (2.3.12a),
    [tex]
    Na^{\dagger} \mid n \rangle = (n+1) a ^{\dagger} \mid n \rangle
    [/tex], I took
    [tex]
    a^{\dagger} \mid n \rangle = c \mid n + 1 \rangle
    [/tex], where [itex]c[/itex] is some constant. Then,
    [tex]
    \langle n \mid a a^{\dagger} \mid n \rangle = | c |^2
    [/tex]
    To find [itex]a^{\dagger} a[/itex],
    [tex]
    \begin{eqnarray}
    a a^{\dagger}
    & = & \left( \frac{m \omega}{2 \hbar} \right) \left( x + \frac{ip}{m \omega} \right) \left( x - \frac{ip}{m \omega} \right)\\
    & = & \frac{H}{\hbar \omega} + \frac{1}{2}\\
    \end{eqnarray}
    [/tex]
    Since it is proven in equation (2.3.5) and (2.3.6) that [itex]H = \hbar \omega (N + \frac{1}{2})[/itex], I get,
    [tex]
    a a^{\dagger} = N + 1
    [/tex]
    Then,
    [tex]
    \langle n \mid a a^{\dagger} \mid n \rangle = \langle n \mid N + 1 \mid n \rangle = n + 1
    [/tex]
    Therefore, from above,
    [tex]
    c = \sqrt{n+1}
    [/tex], which gives,
    [tex]
    a^{\dagger} \mid n \rangle = \sqrt{n+1} \mid n + 1 \rangle
    [/tex], which is the desired result.

    But is there an easier way to get this?

    At the same time,
    [tex]
    a a^{\dagger} = N^{\dagger}
    [/tex]
    So,
    [tex]
    \langle n \mid a a^{\dagger} \mid n \rangle = \langle n \mid N^{\dagger} \mid n \rangle = n^{*}
    [/tex]
    Therefore,
    [tex]
    n+1 = n^{*}
    [/tex]
    How can this happen? I can't see this happening unless the real component of [itex]n[/itex] tends to be infinity.
     
    Last edited: May 1, 2013
  2. jcsd
  3. May 1, 2013 #2
    [itex] N^\dagger = (a^\dagger a)^\dagger = a^\dagger (a^\dagger)^\dagger = a^\dagger a \neq a a^\dagger[/itex]

    Hope that helps :)
     
  4. May 1, 2013 #3
    LOL, thanks.
     
  5. May 1, 2013 #4
    The commutation relation
    [tex]\left[\hat{a},\hat{a}^{\dagger}\right] = 1[/tex]
    will be very, very useful for the SHM and many other applications involving similar-looking Hamiltonians.
     
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